1. rates of change

Hey all, I've been stuck on these two questions for a very long time. I'm stumped on this. And I only got 3 more hours to do this. I'm really screwed, this is the last set of problems and I can't seem to solve them. I also posted in the other section (college/university) section. Please trash that. I didn't know about this section, didn't appear on the top. Stupid loading.

1. A machine is rolling a metal cylinder under pressure. the radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume F is 128pi cubic inches at what rate is the length H changing when the radius r is 1.8 inches?
A. 2.195 in/sec
b. None of these
C. 39.51 in/sec

2. A trough is 15 ft long and 2.4 ft across the top. its ends are isosceles triangles with an altitude of 1.7 ft and vertex down. Water is being pumped into the trough at a rate of 2.3 ft^3/min. how fast is the water level rising when the water is 1.2 ft deep?

Thanks for helping! Btw, great site, will defnetly contribute!

2. Originally Posted by snackxmachine

1. A machine is rolling a metal cylinder under pressure. the radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume F is 128pi cubic inches at what rate is the length H changing when the radius r is 1.8 inches?
A. 2.195 in/sec
b. None of these
C. 39.51 in/sec
You are supposed to assume the the volume of the cylinder is a constant.
Then the volume is:

$V=\pi\,r^2\, h$,

where $r$ is the radius, and $h$ is the length.

Rearrange to:

$h=\frac{V}{\pi\,r^2}$

then differentiate wrt time:

$\frac{dh}{dt}=-\frac{2\,V}{\pi \,r^3}\,\frac{dr}{dt}=-\frac{2\times 128\times\pi}{\pi \times 1.8^3}\times (-0.05) \approx 2.195 \mbox{ inches/s }$

RonL

3. Originally Posted by snackxmachine
2. A trough is 15 ft long and 2.4 ft across the top. its ends are isosceles triangles with an altitude of 1.7 ft and vertex down. Water is being pumped into the trough at a rate of 2.3 ft^3/min. how fast is the water level rising when the water is 1.2 ft deep?
When the depth of water is $h$ the width of the surface is $\frac{2.4}{1.7}h$ (the crossection of the filled part is similar to the complete crossection of the trough). So the volume of water in the trough when filled to depth $h$ is:

$V=\frac{\frac{2.4}{1.7}h\,h}{2}\approx 0.706 h^2$

Differentiate wrt time:

$\frac{dV}{dt}=1.412 h \frac{dh}{dt}$

But $dV/dt=2.3 \mbox{ cubic ft/min}$ so when $h=1.2 \mbox{ ft}$:

$\frac{dh}{dt}=(2.3/1.412)/h=(2.3/1.412)/1.2=1.357 \mbox{ ft/min}$

RonL